Question

In a $△ABC,$ if $\sin A=x,\cos B=y,$ then ${\cos }^{2}C=x^2-2xy\sin C+y^2$.

• A. True
• B. False

Answer 1

The correct option is A True

Given$:\sin A=x\Rightarrow \cos A=\pm \sqrt{1-x^2}$
and $\cos B=y\Rightarrow \sin B=\pm \sqrt{1-y^2}$
We have $A+B+C=\pi$
$\Rightarrow \sin C=\sin (\pi -(A+B))$
$=\sin (A+B)$
$=\sin A\cos B+\cos A\sin B$
$=xy\pm \sqrt{1-x^2}\sqrt{1-y^2}$
$=xy\pm \sqrt{(1-x^2)(1-y^2)}$
and $\cos C=\cos (\pi -(A+B))$
$=-\cos (A+B)$
$=-[\cos A\cos B-\sin A\sin B]$
$=\sin A\sin B-\cos A\cos B$
$=\pm x\sqrt{1-x^2}-(\pm y\sqrt{1-x^2})$
$=\pm x\sqrt{1-x^2}\pm y\sqrt{1-x^2}$
${\cos }^{2}C=x^2(1-y^2)+y^2(1-x^2)\pm 2xy\sqrt{(1-x^2)(1-y^2)}$
$=x^2+y^2-2x^2{y}^2\pm 2xy\sqrt{(1-x^2)(1-y^2)}$
$=x^2+y^2-2xy[xy\pm \sqrt{(1-x^2)(1-y^2)}]$
$=x^2+y^2-2xy\sin C$ where $\sin C=xy\pm \sqrt{(1-x^2)(1-y^2)}$ from above